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give the result for the other three. The exponent in the ﬁrst integrand is x2
x , 2T + x , 1 2 T
2 = , 21 x , T , T 2 + 1 2T + T
2
T 2
T
= , 21 x , rT , 2 + rT:
T
In the ﬁrst integral we make the change of variable p p y = x , rT= , T=2= T; dy = dx= T;
to obtain ~
e,rT S 0 Z m exp
p
2T ~
b x2
x , 2T + x , 1 2 T dx
2
~
1 S 0 Z m exp , 1 x , rT , T 2 dx
=p
~
2T
2
2T
b
p
m p
p~T , r T , 2 T = p 1 S 0:
2T
b " Z p
p
~
pT , r T , 2 T p 2 expf, y2 g dy p ! p p ! ~
m
~
= S 0 N p , r T , 2 T , N pb , r T , 2 T
T
T
Putting all four integrals together, we have " p p ! p : p ! ~
m
~
v 0; S 0 = S 0 N p , r T , 2 T , N pb , r T , 2 T
" T
p
p ! T ~
p
p !
m
~
b
,rT K N p , r T + T , N p , r T + T
,e
2
2
" T p
p ! T~ p
p !
m + r T + T , N 2m , b + r T + T
~
~
p
, S 0 N pT
2
2
T
p !
p m r T
~
+ exp ,rT + 2m r , 2
~
N p +
, 2T ,
p
p ! T
~ ~
p
N 2m , b + r T , 2 T ;
T
where ~ = 1 log K ; m = 1 log L :
b
S 0 ~
S 0 ...
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 Fall '15
 mr.somebody

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